基于图论的奖金分配问题

五、详细设计
#include <iostream>
#include <vector>
using namespace std;
//图结点
typedef struct Node
{
int to;
Node* next;
Node(int to_ = 0, Node* next_ = NULL)
{
to = to_;
next = next_;
}
}Node;
//判断图是否有环
bool isCircle(const vector<Node> &edges) {
int i;
//记录各个结点的入度
vector<int> indegree(edges.size(), 0);
for (i = 1; i < edges.size(); ++i)
{
Node* node = edges[i].next;
while (node)
{
indegree[node->to] ++;
node = node->next;
}
}
//记录入度为0的结点
vector<int> indegree0;
for (i = 1; i < edges.size(); ++i)
{
if (indegree[i] == 0)
{
indegree0.push_back(i);
}
}
//利用队列实现是否存在环的判断
for (i = 0; i < indegree0.size(); ++i)
{
Node *node = edges[indegree0[i]].next;
while (node)
{
indegree[node->to] --;
if (indegree[node->to] == 0)
{
indegree0.push_back(node->to);
}
node = node->next;
}
}
return indegree0.size() < edges.size() - 1; }
//利用bfs计算各个员工的奖金
int solve(const vector<Node> &edges)
{
int i;
//记录各个结点的入度
vector<int> indegree(edges.size(), 0);
for (i = 1; i < edges.size(); ++i)
{
Node* node = edges[i].next;
while (node)
{
indegree[node->to] ++;
node = node->next;
}
}
//记录入度为0的结点
vector<int> tmp;
for (i = 1; i < edges.size(); ++i)
{
if (indegree[i] == 0)
{
tmp.push_back(i);
}
}
vector<bool> isVisited(edges.size(), false);
vector<int> bnous(edges.size(), 100);
//利用队列实现bfs，利用bfs计算各个员工的奖金for (i = 0; i < tmp.size(); ++i )
{
Node* node = edges[tmp[i]].next;
while (node)
{
if (bnous[node->to] < bnous[tmp[i]] + 1)
{
bnous[node->to] = bnous[tmp[i]] + 1;
}
tmp.push_back(node->to);
node = node->next;
}
}
int ret = 0;
//输出奖金
for (i = 1; i < bnous.size(); ++i)
{
ret += bnous[i];
}
return ret;
}
//主函数
int main()
{
vector<Node> edges;
int n, m;
cin>>n>>m;
edges.resize(n + 1);
//输入边
while (m--)
{
int from, to;
cin>>to>>from;
//构造图
Node *newNode = new Node(to, edges[from].next);
edges[from].next = newNode;
}
//判断是否存在环
if (isCircle(edges))
{
cout<<"Poor Xed"<<endl;
}
else
{
cout<<solve(edges)<<endl;
}
return 0;
}。